Solve i(3+i)(2-i)+(2+i)^2-3i

$$i ( 3 + i ) ( 2 - i ) + ( 2 + i ) ^ { 2 } - 3 i$$

Evaluate

$$4+8i$$

Solution Steps

Multiply i and 3+i to get -1+3i.

$$\left(-1+3i\right)\left(2-i\right)+\left(2+i\right)^{2}-3i$$

Multiply -1+3i and 2-i to get 1+7i.

$$1+7i+\left(2+i\right)^{2}-3i$$

Do the additions.

$$\left(2+i\right)^{2}+1+4i$$

Calculate 2+i to the power of 2 and get 3+4i.

$$3+4i+1+4i$$

Do the additions.

$$4+8i$$

Real Part

$$4$$

Solution Steps

Multiply i and 3+i to get -1+3i.

$$Re(\left(-1+3i\right)\left(2-i\right)+\left(2+i\right)^{2}-3i)$$

Multiply -1+3i and 2-i to get 1+7i.

$$Re(1+7i+\left(2+i\right)^{2}-3i)$$

Do the additions in 1+7i-3i.

$$Re(\left(2+i\right)^{2}+1+4i)$$

Calculate 2+i to the power of 2 and get 3+4i.

$$Re(3+4i+1+4i)$$

Do the additions in 3+4i+1+4i.

$$Re(4+8i)$$

The real part of 4+8i is 4.

$$4$$